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6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 1 of 40
6.013, Electromagnetics and ApplicationsProf. Markus Zahn
September 27 and 29, 2005Lectures 6 and 7: Polarization, Conduction, and Magnetization
I. Experimental Observation: Dielectric Media
A. Fixed Voltage - Switch Closed ov = V
As an insulating material enters a free-space capacitor at constant voltagemore charge flows onto the electrodes; i.e., as x increases, i increases.
B. Fixed Charge - Switch open (i=0)
As an insulating material enters a free space capacitor at constant charge,the voltage decreases; i.e., as x increases, v decreases.
II. Dipole Model of Polarization
A. Polarization Vector P Np N q d ( p q d dipole moment)
N dipoles/Volume (P is dipole density) d q
q
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 2 of 40
Courtesy of Krieger Publishing. Used with permission.
inside V pS V
Q = qN d da dV
paired charge or equivalentlypolarization charge density
6.013, Electromagnetic FProf. Markus Zahn
inside V PS V V
Q = P da P dV dV (Divergence Theorem)
P = qN d Np
PP
B. Gauss’ Law
o total u P uE P
o E
oD E
uD
C. Boundary
Unpaired chargedensity; alsocalled free charge
ields, Forces, and Motion Lectures 6 & 7Page 3 of 40
uP
P Displacement Flux Density
Conditions
density
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 4 of 40
a bu u suS V
D = D da = dV n D D
a bP P spS V
P = P da = dV n P P
a bo u P o u P o su spS V
E E da dV n E E =
D. Polarization Current Density
Q qN dV qN d da P da [Amount of Charge passing through
surface area element da ]
p
Q Pdi da
t t[Current passing through surface
area element da]
pJ da
polarization current density
pP
Jt
Ampere’s law:
u p oEx H J Jt
u oP EJt t
u oJ E Pt
u 0DJ ; D E Pt
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 5 of 40
III. Equipotential Sphere in a Uniform Electric Field
o o orlim r, E r cos E z E r cos
r R, 0
3
o 2
Rr, E r cos
r
This solution is composed of the superposition of a uniform electric fieldplus the field due to a point electric dipole at the center of the sphere:
dipole 2o
p cos4 r
with 3o op 4 E R
This dipole is due to the surface charge distribution on the sphere.
3
s o r o o o 3r R r R
2Rr R, E r R, E 1 cos
r r
o o3 E cos
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 6 of 40
IV. Artificial Dielectric
sv v
E , Ed d
sA
q A vd
q A
Cv d
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
For spherical array of non-interacting spheres (s >> R)
z
_3 3
o o z z o oP = 4 R E i P = N p = 4 R E N
31N = s
3 3
o e o e
R RP = 4 E = E = 4
s s
e (electric susceptibility)
o o eD = E P = 1 E = E
r(relative dielectric constant)
3
r o o e o
R= = 1 = 1 4
s
V. Demonstration: Artificial Dielectric
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 7 of 40
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 8 of 40
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
s
v vE = = E =
d d
s
A q Aq = A = v C = =
d v d
o
s
vi = C V =
R
3o
o
A R AC = = 4
d s d
R=1.87 cm, s=8 cm, A= (0.4)2 m2, d=0.15m
=2(250 Hz), Rs=100 k , V=566 volts peak
C=1.5 pf
0 sv = C R V
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 9 of 40
=(2) (250) (1.5 x 10-12) (105) 566 = 0.135 volts
VI. Plasma Conduction Model (Classical)
pdvm = q E m v
dt n
pdvm = q E m v
dt n
p = n kT , p = n kT
k=1.38 x 10-23 joules/oK Boltzmann Constant
A. London Model of Superconductivity [ T 0 , 0 ]
dv dvm = q E , m = q Edt dt
J = q n v , J = q n v
2q E q ndJ d dvq n v = q n = q n = E
dt dt dt m m
2p
2q E q ndJ d dv= q n v = q n = q n = E
dt dt dt m m
2p
2 22 2p p
q n q n= , =
m m(p = plasma frequency)
For electrons: q-=1.6 x 10-19 Coulombs, m-=9.1 x 10-31 kg
n-=1020/m3 , 12o 8.854 x10 farads/m
211
p
q n= 5.6 x10
mrad/s
6.013, Electromagnetic Fields, Forces, aProf. Markus Zahn
p 10
pf = 9 x102
Hz
B. Drift-Diffusion Conduction [Neglect inertia]
0
n k Tdv q k T
m = q E m v V = E ndt n m m n
0
n k Tdv q k T
m = q E m v v = E ndt n m m n
2q n q k TJ = q n v = E n
m m
2q n q k TJ = q n v = E n
m m
= q n , = q n
J = E D
J = E D
q k T= , D =
m m
q k T= , D =
m m
charge mobilities
D D k T= =q
= ther
Einstein’s Relation
MolecularDiffusion
nd Motion Lectures 6 & 7Page 10 of 40
mal voltage (25 mV@ T300o K)
Coefficients
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 11 of 40
C. Drift-Diffusion Conduction Equilibrium J J 0
J = 0 = E D = D
J = 0 = E D = D
D k T= = ln
q
D k T
= = lnq
q / kTo= e
Boltzmann Distributionsq / kT
o= e
o= 0 = = 0 = [Potential is zero when system is charge neutral]
2 q /kT q /kTo o2 q= = = e e = sinhkT
(Poisson - Boltzmann Equation)
Small Potential Approximation:q
1kT
q qsinh
kT kT
2 02 q
= 0k T
2d2
od
k T= 0 ; L =2 qL
Debye Length
6.013, Electromagnetic Fields, Forces, and MotionProf. Markus Zahn
D. Case Studies
1. Planar Sheet
d d
2x /L x /L
1 22 2d
d= 0 = A e A e
dx L
B.C. x = 0
ox 0 = V
d
d
x /Lo
x /Lo
V e x 0
x
V e x 0
/0
dx LV e
0V
( )
Lectures 6 & 7Page 12 of 40
/0
dx LV e
6.013, Electromagnetic Fields, Forces, and MotionProf. Markus Zahn
d
d
x /Lo
d
x
x /Lo
d
V e x 0L
dE = =
dxV
e x 0L
d
d
x /Lo2
d
x
x /Lo2
d
Ve x 0
L
dE= =
dxV
e x 0L
o
s x xd
2 Vx = 0 = E x 0 E x 0 =
L
2. Point Charge (Debye Shielding)
22
d
= 0L
22
1 rr rr
2
2
1= rr r
E. Ohmic Conduction
J = E D
J = E D
If charge density gradients smal
oJ = J J = E =
= ohmJ = E (Ohm’s Law)
Lectures 6 & 7Page 13 of 40
l, then negligible o= =
E = E
ic conductivity
d
d
2
2 2d
r /Lr /Ld1 2
r /L
d rr = 0 0
dr L
r = A e A e
Qr = e
4 r
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 14 of 40
F. p—n Junction Diode
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 15 of 40
A Dn p 2
i
N Nk T= = ln
q n
2 2A p D n
p n
qN x qN xx = 0 = =
2 2
22A pD n
n p
qN xqN x= = +
2 2
D nn p
qN x= x x
2
VII. Relationship Between Resistance and Capacitance In Uniform Media Describedby and .
u S S
L L
D da E daq
C = = =v E ds E ds
L L
S S
E ds E dsvR = = =i J da E da
v
6.013, Electromagnetic Fields, Forces,Prof. Markus Zahn
L L
S L
E ds E daRC = =
E da E ds
Check:
Parallel Plate Electrodes: l AR = , C = RC =
A l
Coaxial
bln aR = , C =b2 l ln
2
Concentric Spherica
and M
l
a
l
,
otion Lectures 6 & 7Page 16 of 40
RC =
,
6.013, Electromagnetic Fields, ForcesProf. Markus Zahn
1 2
1 1R R
R = , C =4
VIII. Charge Relaxation in U
uuJ = 0
t
uE =
uJ = E
uE
t
u
u u
e
=t
IX. Demonstration 7.7.1 –
,
, and Motion Lectures 6 & 7Page 17 of 40
1 2
RC =1 1R R
4
niform Conductors
u
u= 0 = 0t
e = = dielectric relaxation time
etu 00 = r, t = 0 e
Relaxation of Charge on Particle in Ohmic Conductor
6.013, Electromagnetic Fields, Forces, and MoProf. Markus Zahn
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
u
S S
qJ da = E da = =
+
+
+
++
++ +
,
++
++
tion
dqdt
Lectures 6 & 7Page 18 of 40
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 19 of 40
ete
e
dq q= 0 q = q t = 0 e =
dt
Partially Uniformly Charged Sphere
Courtesy of Krieger Publishing. Used with permission.
0 1
3u 1 0
1
r R
4t = 0 = Q = R
30 r R
et0 1 e
u
1
e r R =t =
0 r R
e e
e
t t0
131
r
t
1 22
220
r e Q r e= 0 r R3 4 R
E r, t =
Q eR r R
4 rQ
r R4 r
su 2 0 r 2 r 2r = R = E r = R E r = R
et2
2
Q= 1 e
4 R
Figure 3-21 An initial volume charge distribution within an Ohmic conductor decays exponentiallytowards zero with relaxation time =/and appears as a surface charge at an interface of discontinuity.Initially uncharged regions are always uncharged with the charge transported through by the current.
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 20 of 40
X. Self-Excited Water Dynamos
A. DC High Voltage Generation (Self-Excited)
As the drops form near the rings, image charges are induced on thesurface of the water. When the drops are completely formed (andinsulated from one another) they carry a net charge, the opposite sign tothe sign of charge on the surrounding ring.
Water drops fall into the cans through cross-connected wire loops. A potential difference of more than20 kV between cans is spontaneously generated by the motion of the drops. For optimum operation thedrops should form nearer the rings than shown. This is accomplished by increasing the flow rate.
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 21 of 40
1
2
st21 2i 1 1 i
st12 1i 2 2 i
dvnC v = C v = V e nC V = C s V
dt
dvnC v = C v = V e nC V = C s V
dt
i1
i2
nC V1Cs
= 0nC
1 VCs
Det = 0
2
i inC nC= 1 s =
Cs C
root blows up
inCt
CeAny perturbation grows exponentially with time
B. AC High Voltage Self – Excited Generation
1
2
3
st2i 1 1
st3i 2 2
st1i 3 3
dvnC v = C ; v = V e
dtdv
nC v = C v = V edtdv
nC v = C v = V edt
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 22 of 40
1i
2i
3i
VnC Cs 0
0 nC Cs V = 0
Cs 0 nC V
det = 0
13 3 i 3
i
nCnC + Cs = 0 s = 1
C
1 is = nC C (exponentially decaying solution)
i2,3
nCs = 1 3 j
2C (blows up exponentially because
sreal >0 ; but alsooscillates at frequency simag 0)
XI. Conservation of Charge Boundary Condition
uuJ = 0
t
1 3 1 3j1 1,
2
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 23 of 40
u uS V
dJ da dV = 0
dt
a b su
dn J J = 0
dt
XII. Maxwell Capacitor
A. General Equations
x
x
_
a
_
b
E t i 0 x a
E =
E t i b x 0
a
x b ab
E dx = v t = E t b E t a
sua b a a b b a a b b
d dn J J = 0 E t E t E t E t = 0dt dt
b a
v t aE t = E tb b
b ba a a a a a
dE t v(t) E t a E t v(t) E t a = 0
b dt b
bb a b ba a a
v ta dE a dvE t =
b dt b b b dt
v(t)a, a
b, b
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 24 of 40
B. Step Voltage: v t = V u t
Then dv= V t
dt (an impulse)
At t=0
b a b ba
a dE dv= = V tb dt b dt b
Integrate from t=0- to t=0+
+t 0 0t 0
b a b b ba a a
t 0t 0 t 0
a dE adt = E = V t dt = V
b dt b b b
aE t 0 = 0
b b b
a a aa b
a VE t = 0 = V E t = 0 =
b b b a
For t > 0, dv= 0
dt
b a b ba a a
a dE aE t = V
b dt b b
tb a b
aa b a b
V b aE t = A e ; =
b a b a
b b b ba
a b a b a b a b
V VE t = 0 = A = A = V
b a b a b a b a
t tb b
aa b a b
V VE t = 1 e e
b a b a
b aV aE t = E tb b
V
v(t)
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 25 of 40
su a a b b a a b a
V at = E t E t = E t E t
b b
ba a b
a V= E t
b b
b a a b t
a b
V= 1 e
b a
C. Sinusoidal Steady State:
j tv t = Re V e
j ta aE t = Re E e
b
j tbE t = Re E e
Conservation of Charge Interfacial Boundary Condition
a a b b a a b b
dE t E t E t E t = 0
dt
a a a b b bˆ ˆE j E j = 0
b aˆ ˆE b E a = V
ab
ˆˆ E aVE =
b b
a
a a a b b
ˆˆ E aVE j j = 0
b b
a a a b b b b
ˆa VE j j = j
b b
a b
b b a a a a b b
ˆ ˆ ˆE E V= =
j j b j a j
su a a b bˆ ˆ= E Eˆ
a b b a
a a b b
= Vb j a j
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 26 of 40
D. Equivalent Circuit (Electrode Area A)
a a a b b b
ˆ ˆI = j E A = j E A
a b
a a b b
V=
R RR C j 1 R C j 1
a ba b
a bR = , R =
A A
a ba b
A AC = , C =
a b
v(t)
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 27 of 40
XIII. Magnetic Dipoles
Courtesy of Krieger Publishing. Used with permission.
Courtesy of Krieger Publishing. Used with permission.
6.013, Electromagnetic Fields, Forces, andProf. Markus Zahn
Diamagnetism
2z
e e eI = = , m = I R i =
2 2 2
2_ _2
z ze R
R i = i2
Angular Momentum _ _ _ _
2r r ze e eL = m R i v = m R R i i = m R i
r p e2m= m
e
linear momentum
L is quantized in units of 34h , h = 6.62x10 joule sec2
(Planck’s constant)
24 2
e e e
e L e h ehm = = = 9.3x10 amp m
2m 2 2 m 4 m
Bohr magneton mB
(smallest unit ofmagnetic moment)
Imagine all Bohr magnetons in sphere of radius R aligned. Net magnetic moment is
3 0B
0
A4m m R3 M
Total massof sphere
For iron: =7.86 x 103 kg/
Courtesy of Hermann A.
Motion Lectures 6 & 7Page 28 of 40
molecular weight
m3, M0=56
Haus and James R. Melcher. Used with permission.
Avogadro’s number = 6.023 x 1026 molecules per kilogrammole
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 29 of 40
For a current loop
2 3 0 0B B
0 0
A A4 4m = i R = m R i = m R3 M 3 M
For R = 10 cm
26
-24 36.023 104i = 9.3 10 .1 7.86 10
3 56
= 1.05 x 105 Amperes
Thus, an ordinary piece of iron can have the same magnetic moment as acurrent loop of radius 10 cm of 105 Amperes current.
B. Magnetic Dipole Field
_ _0
r30
mH = 2 cos i sin i4 r
(multiply top & bottom by 0 )
Electric Dipole Field
_ _
r30
pE = 2 cos i sin i4 r
Analogy
0p m
P = Np M = Nm , N = # of magnetic dipoles / volume
Polarization Magnetization
6.013, Electromagnetic Fields, Forces, and Motion Lectures 6 & 7Prof. Markus Zahn Page 30 of 40
XIV. Maxwell’s EQS Equations with Magnetization
A. Analogy to Maxwell’s EQS Equations with Polarization
EQS
0 uE = P
p = P (Polarization or paired
charge density)
α b a b
0 sun E E = n P P
a b
sp = n P P
MQS
0 0H = M
m 0= M (magnetic charge
density)
α b a b
0 0n H H = n M M
a b
sm 0= n M M
H = J
0E = H Mt
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 31 of 40
B. MQS Equations
0B = H M Magnetic flux density B has units of Teslas
(1 Tesla = 10,000 Gauss)
B 0
a an B B = 0
BE =
t
H = J
S
dv = , = B da
dt(total flux)
XV. Magnetic Field Intensity along Axis of a Uniformly Magnetized Cylinder
a b
sm 0 sm 0 0d= n M M z = = M2
sm 0 0dz = = M2
x H = J = 0 H =
6.013, Electromagnetic Fields, Forces, and MotionProf. Markus Zahn
20 0 m 0H = = = M
m2 m
0 V' 0
r ' dV '= r =
4 r r '
R Rsm sm
r'=0 r'=00 0
d dz = 2 r 'dr ' z = 2 r 'dr '2 2z =4 r r ' 4 r r '
R R
0 o 0 o1 1
2 22 2r'=0 r'=02 2
0 0
M 2 r 'dr ' M 2 r 'dr '=
d d4 r ' z 4 r ' z2 2
1
2 221
2 22
r 'dr ' = r ' z ar ' z a
RR1 12 22 2
2 20
r ' 0 r ' 0
M d dz = r ' + z r ' z
2 2 2
1 12 22 2
2 20M d d d d= R z z R z z
2 2 2 2 2
( )z
Lecture 6 & 7Page 32 of 40
2 2 1/ 20( / 2){ [ ] }M R d R d
6.013, Electromagnetic Fields, Forces, and MotionProf. Markus Zahn
01 1
2 22 22 2
dd zzM d22 z2 2
d dR z R z2 2
zH = =z
01 1
2 22 22 2
dd zzM d d22 2 z2 2 2d dR z R z
2 2
XVI. Toroidal Coil
Courtesy of Hermann A. Haus and James R. M
-d/2 d/
Hz
Lecture 6 & 7Page 33 of 40
elcher. Used with permission.
2
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 34 of 40
1 1
1C
N i N iH dl = H 2 r = N i H =
2 r 2 R
2wB
4
2
2 2w
= N = N B4
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
H 1 1 1 H
1
H 2 RV = i R = R V =Horizontalvoltage to oscilloscope
N
2 v2 2 2 v v 2 2
d dVv = = i R V = V R Cdt dt
If 2 v2 2 2 2 2 2 v v
2
d dV1R R C = R C V V =Verticalvoltage tooscilloscopeC dt dt
2
2
w= N B
4
2
v 22 2
1 wV = N B
R C 4
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 35 of 40
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 36 of 40
XVII. Magnetic Circuits
In iron core:H = 0
lim B = H
B finite
NiH dl = Hs = Ni H =s
0
0
Dd N= H Dd = i
s
S
B da = 0
2 20 0Dd Dd= N = N i L = = N
s i s
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 37 of 40
XVIII. Reluctance
0
lengthNi s= = =Dd permeability cross sec tionalarea
R
[Reluctance, analogous to resistance]
Series
Parallel
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 38 of 40
A. Reluctances In Series
1 2
1 21 1 2 2
s s= , =
a D a DR R
1 2
Ni=R R
1 1 2 2C
H dl = H s H s = Ni
1 1 1 2 2 2= H a D = H a D
2 2 1 11 2
1 1 2 2 2 1 1 1 2 2 2 1
a Ni a NiH = ; H =
a s a s a s a s
B. Reluctances In Parallel
1 2 1 2C
NiH dl = H s = H s = Ni H = H
s
1 21 1 1 2 2 2 1 2
1 2
Ni= H a H a D = = Ni
R RP P
R R
1 21 2
1 1= ; =P P
R R
1=P
R[Permeances, analogous to Conductance]
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 39 of 40
XIX. Transformers(Ideal)
Courtesy of Krieger Publishing. Used with permission.
A. Voltage/Current Relationships
1 1 2 2N i N i l= ; =
AR
R
Another way: 1 1 2 2C
H dl = Hl = N i N i
1 1 2 2N i N iH =l
1 1 2 21 1 2 2
N i N iA= HA = N i N i =
l R
6.013, Electromagnetic Fields, Forces, and Motion Lecture 6 & 7Prof. Markus Zahn Page 40 of 40
21 1 1 1 1 2 2 1 1 2
A= N = N i N N i = L i Mi
l
22 2 1 2 1 2 2 1 2 2
A= N = N N i N i = Mi L i
l
2 21 1 0 2 2 0 1 2 0 0
A 1L = N L , L = N L , M = N N L , L = =l R
12
1 2M = L L
1 1 2 1 21 1 1 0 1 2
d di di di div = = L M = N L N N
dt dt dt dt dt
2 1 2 1 22 2 2 0 1 2
d di di di div = = M L = N L N N
dt dt dt dt dt
1 1
2 2
v N=v N
1 2
1 1 2 22 1
i Nlim H 0 N i = N i =
i N
1 1
2 2
v i= 1
v i
Courtesy of Hermann A. Haus andJames R. Melcher. Used with permission.